Consider the following compound (X) beginarrayc mathrm I \\ mathrm H - mathrm C equiv mathrm C - mathrm C H _ 2 - mathrm C H - mathrm C H _ 3 \\ mathrm I \\ mathrm C H _ 3 endarray$$\begin{array}{c} \mathrm {I} \\ \mathrm {H} - \mathrm {C} \equiv \mathrm {C} - \mathrm {C H} _ {2} - \mathrm {C H} - \mathrm {C H} _ {3} \\ \mathrm {I} \\ \mathrm {C H} _ {3} \end{array}$$ The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding mathrmC - H$\mathrm{C - H}$ bond are :

### Related Formula textBasicity propto textAvailability of lone pair of electrons on Nitrogen$$\text{Basicity} \propto \text{Availability of lone pair of electrons on Nitrogen}$$ ### Core Logic Analyzing the molecules: - In molecule **(R)**, according to Bredt's rule, the bridgehead nitrogen has a localized lone pair which cannot participate in resonance. Thus, it is highly available and most basic. - In molecule **(Q)**, the nitrogen lone pair is involved in cross-conjugation with the carbonyl group, reducing its availability. - In molecule **(P)**, the lone pair on nitrogen is directly conjugated with the carbonyl group (amide resonance), making it the least available. Therefore, the correct basicity order is: R > Q > P$R > Q > P$ ### Step 1: Final Identification

Comparing availability, structure R has localized electrons, Q has cross-conjugation, and P has standard amide resonance. Hence, option (4) is correct. ### Pattern Recognition Look for localized vs delocalized lone pairs on nitrogen. Bridgehead nitrogen lone pairs that violate Bredt's rule for double bond formation remain strictly localized, drastically increasing basicity compared to conjugated amides. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Principles of Organic Chemistry

### Related Formula textStability propto textDelocalization of charge via resonance, mesomeric, and back-bonding effects$$\text{Stability} \propto \text{Delocalization of charge via resonance, mesomeric, and back-bonding effects}$$ ### Core Logic Evaluating each specific structural pair: - **Pair (A):** The first carbocation is stabilized by strong +M$+M$ back-bonding from the methoxy (-OMe$-OMe$) oxygen lone pair, making it significantly more stable than the second. - **Pair (B):** The first carbanion is strongly stabilized by the -M$-M$ and -I$-I$ electronic effects of the nitro (-NO_2$-NO_2$) group situated at the ortho position, whereas a carbocation in that spot would be destabilized. Thus, the first ion is more stable. - **Pair (C):** The second cation has extended allylic resonance stabilization, meaning the first is less stable. - **Pair (D):** Cation with -OMe$-OMe$ backbonding is more stable than tertiary aliphatic carbocation, making the first less stable than the second. Thus, only in **(A) & (B)** is the first ion more stable than the second. ### Pattern Recognition Back-bonding from an adjacent oxygen lone pair always triumphs over standard inductive or hyperconjugative alkyl stability templates. For carbanions, ensure electron-withdrawing groups like -NO_2$-NO_2$ match the sign of the charge. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Principles of Organic Chemistry

### Related Formula textPrincipal Functional Group Priority: -OH > textDouble bond (=) ge textTriple bond (equivtext)$$\text{Principal Functional Group Priority: } -OH > \text{Double bond (=)} \ge \text{Triple bond (}\equiv\text{)}$$ ### Core Logic Number the seven-carbon parent chain from the side that gives the principal functional group (-OH$-OH$) and double bond the lowest locants: - Numbering from right-to-left gives locants: alkene at position 1, alcohol at 4, and alkyne at 6. - Numbering from left-to-right gives locants: alkyne at 1, alcohol at 4, alkene at 6. According to IUPAC rules, when choices are symmetric for the principal group, the lower locant is given to the double bond over the triple bond. Hence, right-to-left numbering is correct: overset7textHoverset6textCequiv overset5textC-overset4textCtextH_2-overset3textCtextH(OH)-overset2textCtextH_2-overset1textCtextH=textCH_2$$\overset{7}{\text{H}}\overset{6}{\text{C}}\equiv \overset{5}{\text{C}}-\overset{4}{\text{C}}\text{H}_2-\overset{3}{\text{C}}\text{H(OH)}-\overset{2}{\text{C}}\text{H}_2-\overset{1}{\text{C}}\text{H}=\text{CH}_2$$ This gives: **Hept-1-en-6-yn-4-ol**. ### Pattern Recognition When terminal unsaturations are tied symmetrically (positions 1 and 6), the alkene ('en') takes prefix allocation priority over the alkyne ('yn'). The secondary alcohol suffix '-ol' forms the principal name ending. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Principles of Organic Chemistry

### Core Logic Evaluating standard NCERT laboratory purification matches: - **(A) Aniline from aniline-water mixture:** Aniline is steam volatile and immiscible with water, so it is separated via **Steam distillation (IV)**. - **(B) Glycerol from spent-lye in soap industry:** Glycerol decomposes at or below its boiling point, hence it is separated via **Distillation at reduced pressure (III)**. - **(C) Different fractions of crude oil:** Separated using their small differences in boiling points via **Fractional distillation (II)**. - **(D) Chloroform-Aniline mixture:** Separated due to a substantial boiling point difference via **Simple distillation (I)**. ### Pattern Recognition Match key words directly: Glycerol/spent-lye always links to reduced pressure (vacuum distillation). Crude oil always couples to fractional columns. Aniline + water implies steam injection. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Principles of Organic Chemistry

### Core Logic Analyzing the mechanism: 1. Protonation (H^+$H^+$ attack) on the double bond occurs to generate the most stable intermediate carbocation. 2. The system forms a highly stable **tertiary (3^circ$3^\circ$) carbocation** at the bridge junction ring site. 3. Nucleophilic attack by bromide ion (Br^-$Br^-$) captures this bridge junction center selectively, yielding the major brominated structure designated as Structure (2). ### Step 1: Mechanism Breakdown

Electrophilic addition follows Markovnikov-like stability rules, directing the halide specifically to the bridgehead junction carbocation position. ### Pattern Recognition Whenever adding hydrohalic acids across cyclic non-conjugated dienes or isolated double bonds, always convert the alkene into the most stable, un-strained tertiary carbocation before attaching the nucleophile. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Principles of Organic Chemistry Class 11 Chemistry: Hydrocarbons